Breaking the denominator into primes

For N / 18k to be an integer you must be able to cancel these prime factors in both the numerator and the denominator. Thus, the question becomes how many times are these prime factors present in the numerator?

There are so many twos in the numerator that you will be able to cancel many twos in the denominator. However, let’s concentrate on counting the threes in the numerator, as there are fewer of them:

Now let’s add the number of threes in the numerator. When you do this, you will see there are a total of 14.

Every time there is an 18 (containing 2 threes) in the denominator, this must be balanced by 2 threes in the numerator. As there are 14 threes in the numerator, we can have a maximum of seven 18s in the denominator. The correct answer is therefore 7 (C).

At first glance it’s difficult to work out how many times 18 is present in the numerator. But once you break 18 down into it prime factors, you can quickly work out how many times each of these factors is present.

This is another example of the usefulness of prime factorization. Keep this tool in mind when questions seem to be too difficult to solve in the two minutes you have!

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